Recall from part a {\displaystyle a} that f ′ ( x ) = e x ( x − 2 ) x 3 {\displaystyle f'(x)={\frac {e^{x}(x-2)}{x^{3}}}} . We take the derivative of f ′ {\displaystyle f'} with respect to x {\displaystyle x} using the quotient rule. Let
g ( x ) = e x ( x − 2 ) g ′ ( x ) = e x ( x − 2 ) + e x = e x ( x − 1 ) h ( x ) = x 3 h ′ ( x ) = 3 x 2 {\displaystyle {\begin{aligned}g(x)&=e^{x}(x-2)\quad \quad \quad g'(x)=e^{x}(x-2)+e^{x}=e^{x}(x-1)\\h(x)&=x^{3}\quad \quad \quad \quad \quad \quad h'(x)=3x^{2}\end{aligned}}}
By the quotient rule:
f ″ ( x ) = g ′ ( x ) h ( x ) − g ( x ) h ′ ( x ) ( h ( x ) ) 2 = e x ( x − 1 ) x 3 − 3 e x ( x − 2 ) x 2 x 6 = e x x 2 ( x ( x − 1 ) − 3 ( x − 2 ) ) x 6 = e x ( x 2 − 4 x + 6 ) x 4 {\displaystyle {\begin{aligned}f''(x)&={\frac {g'(x)h(x)-g(x)h'(x)}{(h(x))^{2}}}\\&={\frac {e^{x}(x-1)x^{3}-3e^{x}(x-2)x^{2}}{x^{6}}}\\&={\frac {e^{x}x^{2}(x(x-1)-3(x-2))}{x^{6}}}\\&={\frac {e^{x}(x^{2}-4x+6)}{x^{4}}}\end{aligned}}}
that 
. We take the derivative of 
with respect to 
using the quotient rule. Let
